"glm" will do multinomial logistic regression. However, if J is large,
I doubt if that will do what you want. If it were my problem, I might
feel a need to read the code for "glm" and modify it to do what I want.
Perhaps someone else can suggest something better.
hth. spencer graves
Christ
Dear all,
In Pinheiro and bates'book, we fit an ARMA(1,1) model with:
>fm5Ovar.lme<-update(fm1Ovar.lme,corr=corARMA(p=1,q=1)
But the result is:
Error in "coef<-.corARMA"(*tmp*, value = c(62.3428455940029,
95.0395441938099, :
Coefficient matrix not invertible
There is also problems when
It is one of the latest virus variants. If you have a windows operating system, check
your
virus software for updates. I must have gotten this 10-15 times in the
past 3 days from one list or the other.
--
Mark Hall
Niigata Prefectural Museum of History
<>
__
Hi,
Does anyone out there know how to add a legend when using the R-function
image?
Ie. is there something out there like the S+ function image.legend?
Doug.
Fisheries Research Services,
Marine Laboratory,
Victoria Road,
Torry,
Aberdeen, UK.
Tel. 44 (0) 1224 295314
I am at a loss to undestand this message since I have made no application
to you (nor, for that matter, is there an attached file). Please delete my
address from all your records and advise me that this has been done.
--On 02 June 2003 13:29 +0100 [EMAIL PROTECTED] wrote:
Please see the attach
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On Mon, 02 Jun 2003 12:34:47 +0200, you wrote:
>"Error in var(x, na.rm = na.rm) : missing observations in cov/cor"
>
>Somebody would give me any clue about the origin of this error?
>There is any probability that the origin of the problem is the scarcitie
>of resources?
>( I don´t understand ho
Rafael Bertola wrote:
I write a function to plot some graphs and a gave for it one result of
lm() (reg) and a character (name). In win.metafile("name-%02d.wmf",
pointsize = 14) i want compose de name of file with the string i pass to
the function. How i can make this?
plota.res.grava <- function(
Alicia Lopez wrote:
I would be gratefulf if anybody can help me with this problem. I have not
a experience with R
language, actually this is my first job with it :
I have write some instructions for acomplish a simulation, and I have 36
conditions
with 1000 iterations each one . The program ru
?paste
as in
win.metafile(paste(name, "%02d.wmf", sep="-"), pointsize = 14)
On Mon, 2 Jun 2003, Rafael Bertola wrote:
> I write a function to plot some graphs and a gave for it one result of
> lm() (reg) and a character (name). In win.metafile("name-%02d.wmf",
> pointsize = 14) i want compose
On Mon, 2 Jun 2003, Andreas Christmann wrote:
> >>> 1. RE: Ordinal data - Regression Trees & Proportional Odds
>(Liaw, Andy)
>
> > AFAIK there's no implementation (or description) of tree algorithm
> > that handles ordinal response.
> >
>
> Regression trees with an ordinal response va
Dear R-Users:
Are there any packages or functions in R to estimate a random effects
ordered probit model? What about a dynamic ordered probit? Finally,
random parameters (mixed effects models) for ordered probits?
Thank you,
Ricardo
Ricardo Sabates
Research Officer
Centre for Research on t
I write a function to plot some graphs and a gave for it one result of
lm() (reg) and a character (name). In win.metafile("name-%02d.wmf",
pointsize = 14) i want compose de name of file with the string i pass to
the function. How i can make this?
plota.res.grava <- function(reg,name){
win.me
>>> 1. RE: Ordinal data - Regression Trees & Proportional Odds
(Liaw, Andy)
> AFAIK there's no implementation (or description) of tree algorithm
> that handles ordinal response.
>
Regression trees with an ordinal response variable can be computed with
SPSS Answer Tree 3.0.
Andreas Christma
1. RE: Ordinal data - Regression Trees & Proportional Odds
(Liaw, Andy)
AFAIK there's no implementation (or description) of tree algorithm that
handles ordinal response.
Regression trees with an ordinal response variable can be computed with
SPSS Answer Tree 3.0.
Andreas Christmann
I would be gratefulf if anybody can help me with this problem. I have not
a experience with R
language, actually this is my first job with it :
I have write some instructions for acomplish a simulation, and I have 36
conditions
with 1000 iterations each one . The program runs without problems,
On Mon, 2 Jun 2003 [EMAIL PROTECTED] wrote:
> Obviously I cannot use the underscore "_ " character to name an object
> in R
Actually, you can but it is more trouble than it is worth as you will
always have to quote the name, and often explicitly get it:
> "my_pi" <- pi
> get("my_pi")
[1] 3.141
Hi R lovers
Obviously I cannot use the underscore "_ " character to name an object in R
Is there a special reason for that ?
I want to use it to rename a function
maybe the problem is due to the nature of the object I work on
thanks for any comments on that very little and not very bothering tro
天才 wrote:
Hello ??,
> hello,everyone!
> I have studied in univ. for two years.and my teacher have put out
> some stat. model .and i want to write it to a R package, and take
> it to my future paper.any advise? i learned something about SAS/stat.
SAS/stat, hmmm, not very interesting
For numbers ending in 5 Excel always rounds up and I want to reproduce this in R
rather than use the round function.
I have tried:
x<-as.integer(x*100+0.5)/100
and
x<-floor(x*100+0.5)/100
However, some values of x cause problems e.g
x<-floor(4.145*100+0.5)/100
4.14
I have tried breaking it d
Dear colleagues,
Von Bertalanffy model is commonly adjust to data on fish length (TL) and age (AGE)
TL= Linf*(1-exp(-K*(AGE-t0)). Linf, K and t0 are parameters of the model.
One main goal of the growth study is the comparison of growth parameter estimates
between sexes of the same species, or es
Hello,
You should start creating a separate directory for your package
(e.g. foo). As a minimum you need three things:
1) your R functions (as .R files in directory foo/R)
2) documentation of your functions in .Rd format (look "writing R
extensions" and function prompt(), they must reside in
Your message
To: [EMAIL PROTECTED]
Subject: Re: Submited (004756-3463)
Sent:Mon, 2 Jun 2003 00:48:50 -0700
did not reach the following recipient(s):
[EMAIL PROTECTED] on Mon, 2 Jun 2003 00:48:02 -0700
The recipient name is not recognized
MSEXCH:IMS:Intel:Americas01:FMSMSX0
I want to do a logistic regression analysis, and to compare with, a
discriminant analysis. The mentioned power maps are my exogenous data,
the dependent variable (not mentioned so far) is a diagnosis
(ill/healthy)
thanks for the interest and the help
Christoph
On Sun, 2003-06-01 at 21:01, Spence
hello,everyone!
I have studied in univ. for two years.and my teacher have put out
some stat. model .and i want to write it to a R package, and take
it to my future paper.any advise? i learned something about SAS/stat.
and i have read the tutorial of R.
now i can interactive with R,
Hi,
I've been trying to learn and use R for data analysis, and so far it
has be OK. It is very slow with the aov command. But, other than that
things are generally OK. Anyway, todays problem is one that I have
looked in quite a few places to try to solve but to know avail. I hope
the list c
This is an automated response. SkillSoft Support has
received your message and this issue has been assigned
ticket number : DF/010989
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identify the original issue, and you are not alloc
Your specific example is a scaled beta. Therefore, "2*rbeta(1000, 4,
1)" will generate 1000 random numbers according to that distribution.
You can get the same distribution from "2*qbeta(runif(1000), 4, 1)".
We can generalize this last example to any case of interest. Example:
df4 <- fun
Hi,
Use the Inverse transformation method. See any basic Cbook in simulation
for instance Sheldon Ross's book.
Regards,
Edgar
On Mon, 2 Jun 2003, Fernando Henrique Ferraz Pereira da Rosa wrote:
> Hi, I'd like to know if there's anything in R that could help me do
> that. Let's suppose I have
Hi, I'd like to know if there's anything in R that could help me do
that. Let's suppose I have a density function of a random variable, for example
f(x) = (x^3)/4 0 < x < 2 and I would like to simulate it. For the common
distributions (exponencial, gamma, cauchy) there are the r-functions (rga
What are you trying to do? What I would do with this depends on many
factors.
spencer graves
Christoph Lehmann wrote:
again, under another subject:
sorry, maybe an all too trivial question. But we have power data from J
frequency spectra and to have the same range for the data of all our
subjec
Im tryng to understand an error i get with integrate. this is 1.7.0 on
solaris 2.8.
##i am trying to approximate an integral of this function,
f<-function(b) exp(-(b-mu)^2/(2*tau2))/(p-exp(b))*10^6
##with
tau2 <- .005;mu <- 7.96;p <- 2000
##from -inf to different upper limits. using
integrate(f,
On Sun, Jun 01, 2003 at 06:53:55PM +0200, Uwe Ligges wrote:
>> but I'd like to persp()' colors behave like in image() function!
> That's not easy, because you have to redefine x, y and z values.
>
> Simple example:
>
> x <- y <- 1:2
> z <- matrix(1:4, 2)
> image(x, y, z)# OK, quite nice
Timur Elzhov wrote:
Dear R experts!
I use image() & persp() functions for color plotting z(x,y)-type
graphics. In image() colors correspond to z-values, that's what
I want. OTOH, in persp() the col option means:
col: the color(s) of the surface facets. Transparent colours are
ig
Dear R experts!
I use image() & persp() functions for color plotting z(x,y)-type
graphics. In image() colors correspond to z-values, that's what
I want. OTOH, in persp() the col option means:
col: the color(s) of the surface facets. Transparent colours are
ignored. This is rec
Good afternoon R-masters,
I am with some doubts in the R, see the script below:
m<-c(69.6,67.3,75.6,74.3,64.7,60,65.7,62.5,66.5)
d<-c(11.6,15,17.8,18.3,11.2,11,4.6,5.8,7)
year<-c(1994,1995,1996,1997,1998,1999,2000,2001,2002)
male<-ts(m,start=c(1994))
death<-ts(d,start=c(1994))
data<
On Sat, 31 May 2003, Paul E. Johnson wrote:
> I keep finding myself in a situation where I want to calculate a
> variable name and then use it on the left hand side of an assignment.
> For example
>
> iteration <- 1
> varName <- paste("run",iteration,sep="")
> myList$parse(text=varName) <- aColumn
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